What is the integral from 0 to t of ue^{t-u} ?

The integral from 0 to t of ue^{t-u} is e^t(-te^{-t}-e^{-t}+1)

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integral from 0 to t of ue^{t-u}

\int_{0}^{t} u e^{t - u} d u

e^{t} (- t e^{- t} - e^{- t} + 1)

Solution steps

\int_{0}^{t} u e^{t - u} d u

Simplify

e^{t - u} : e^{t} e^{- u}

= \int_{0}^{t} e^{t} e^{- u} u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{t} \cdot \int_{0}^{t} e^{- u} u d u

Apply Integration By Parts

: [- e^{- u} u - \int - e^{- u} d u]_{0}^{t}

= e^{t} [- e^{- u} u - \int - e^{- u} d u]_{0}^{t}

\int - e^{- u} d u = e^{- u}

= e^{t} [- e^{- u} u - e^{- u}]_{0}^{t}

Compute the boundaries:

- t e^{- t} - e^{- t} + 1

= e^{t} (- t e^{- t} - e^{- t} + 1)

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What is the integral from 0 to t of ue^{t-u} ?
The integral from 0 to t of ue^{t-u} is e^t(-te^{-t}-e^{-t}+1)